Gaseous butane (CH3(CH2)2 (CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O)

Question

What is the theoretical yield of water formed from the reaction of 4.65g of butane and 26.5g of oxygen gas? Round your answer to 3 significant figures.

= _____ g

Answer 1

2 + 13 O₂ ==> 8 CO₂ + 10 H₂O ... balanced equation

Find limiting reactant. An easy way is to simply divide MOLES of each reactant by the corresponding coefficient in the balanced equation. Whichever value comes out less is the limiting reactant.

For C₄H₁₀: 4.65 g C₄H₁₀ x 1 mol / 58.1 g = 0.0800 moles C₄H₁₀ (÷2->0.04)

For O₂: 26.5 g O₂ x 1 mol O₂ / 32 g = 0.828 mol O₂ (÷13->0.06)

Since 0.04 is less than 0.06 C₄H₁₀ is limiting and the 0.0800 moles of C₄H₁₀ will dictate how much H₂O is formed in the reaction.

Theoretical yield of H₂O:

0.0800 mols C₄H₁₀ x 10 mols H₂O / 2 mols C₄H₁₀ x 18 g H₂O/mol = 7.20 g H₂O

(NOTE: the answer comes out the same as that reported by @Joan L, but that's just luck because if O₂ was limiting the previous answer would have been incorrect as it would have used 9 mols O₂ instead of 13 mols O2.)

Answer 2

Marissa,

The first thing you will need to do is to write the BALANCED chemical equation. This is a combustion reaction producing CO₂ and H₂O

Another way to write the formula of butane is C₄H₁₀ [ I got this by adding the # of each atoms in the formula given]

2 C₄H₁₀ + 9 O₂ → 8 CO₂ + 10 H₂O

I suggest if you do not know how I obtain this equation that you review balancing of chemical equations

Because the question gave 2 amounts of reactants we know it is a limiting reagent problem where one of the reactant determines the product. There are 2 ways to work this problem but for me I prefer to

(a) USE EACH REAGENT AND DETERMINE THE AMOUNT OF THE REQUIRED PRODUCT PRODUCED.

(b) THE REAGENT THAT PRODUCES THE LEAST AMOUNT OF PRODUCT IS CALLED THE LIMITING REAGENT AND THE AMOUNT OF THE PRODUCT PRODUCED WILL BE THE THEORETICAL YIELD OF THE REACTION FOR THE CHOSEN PRODUCT[if more than one].

So using stoichiometry {I am going to presume you know what I am referring to here}

g reactant →mol reactant → mol product → g product

g H₂O produced using butane is

4.65 g C₄H₁₀ *(1 mol C₄H₁₀ / 58.1 g C₄H₁₀) * (10 mol H₂O / 2 mol C₄H₁₀) * (18.0 g H₂O / 1 mol H₂O) =7.20 g

g H₂O produced using oxygen is

26.5 g O₂ *(1 mol O₂ / 32.0 g O₂ ) * (10 mol H₂O / 9 mol O₂) * (18.0 g H₂O / 1 mol H₂O) =16.6 g

Based on the explanation given above the answer is 7.20 g H₂O

All the best!